Find $A$ and $B$ such that
\[\frac{3x+5}{x^2-x-42}=\frac{A}{x-7}+\frac{B}{x+6}.\]Write your answer in the form $(A,B)$.
We factor the denominator in the left-hand side to get \[\frac{3x+5}{(x-7)(x+6)}= \frac{A}{x - 7} + \frac{B}{x + 6}.\]We then multiply both sides by $(x - 7)(x + 6)$, to get \[3x + 5 = A(x + 6) + B(x - 7).\]We can solve for $A$ and $B$ by substituting suitable values of $x$.  For example, setting $x = 7$, the equation becomes $26 = 13A$, so $A = 2$.  Setting $x = -6$, the equation becomes $-13 = -13B$, so $B = 1$.  Therefore, $(A,B) = \boxed{(2,1)}$.